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Two moles of ideal helium gas are in a r...

Two moles of ideal helium gas are in a rubber balloon at `30^@C.` The balloon is fully expandable and can be assumed to require no energy in its expansion. The temperature of the gas in the balloon is slowly changed to `35^@C.` The amount of heat required in raising the temperature is nearly (take R
`=8.31 J//mol.K`)

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To solve the problem of calculating the amount of heat required to raise the temperature of two moles of ideal helium gas in a rubber balloon from \(30^\circ C\) to \(35^\circ C\), we can follow these steps: ### Step 1: Identify the parameters - Number of moles, \( n = 2 \) moles - Initial temperature, \( T_i = 30^\circ C = 30 + 273.15 = 303.15 \, K \) - Final temperature, \( T_f = 35^\circ C = 35 + 273.15 = 308.15 \, K \) - Change in temperature, \( \Delta T = T_f - T_i = 35 - 30 = 5^\circ C \) ### Step 2: Determine the specific heat capacity at constant pressure, \( C_p \) For a monatomic ideal gas like helium, the specific heat capacity at constant pressure is given by: \[ C_p = \frac{5}{2} R \] Where \( R = 8.31 \, J/(mol \cdot K) \). ### Step 3: Calculate \( C_p \) Substituting the value of \( R \): \[ C_p = \frac{5}{2} \times 8.31 = 20.775 \, J/(mol \cdot K) \] ### Step 4: Calculate the heat required, \( Q \) The heat required at constant pressure is calculated using the formula: \[ Q = n C_p \Delta T \] Substituting the values: \[ Q = 2 \times 20.775 \times 5 \] ### Step 5: Perform the calculation Calculating \( Q \): \[ Q = 2 \times 20.775 \times 5 = 207.75 \, J \] ### Step 6: Round off the answer Rounding \( 207.75 \, J \) gives us approximately: \[ Q \approx 208 \, J \] ### Final Answer The amount of heat required to raise the temperature is nearly **208 Joules**. ---

To solve the problem of calculating the amount of heat required to raise the temperature of two moles of ideal helium gas in a rubber balloon from \(30^\circ C\) to \(35^\circ C\), we can follow these steps: ### Step 1: Identify the parameters - Number of moles, \( n = 2 \) moles - Initial temperature, \( T_i = 30^\circ C = 30 + 273.15 = 303.15 \, K \) - Final temperature, \( T_f = 35^\circ C = 35 + 273.15 = 308.15 \, K \) - Change in temperature, \( \Delta T = T_f - T_i = 35 - 30 = 5^\circ C \) ...
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