`5g` of a gas is contained in a rigid container and is heated from `15^@C to 25^@C` . Specific heat capacity of the gas at constant volume is `0.172 cal g ^(-1)^@C^(-1)` and the mechanical equivalent of heat is 4.2 J cal . Calculate the change in the internal energy of the gas.

Explanation

`m = 5 g`,
`Delta = 25 -15^@C = 10^@ C`
`C_v = 0.172cal//g -^@C,`
`J = 4.2J//cal`.
We know,
`dQ = dU+dW`
Now, `V=0`( for a rigid body)
So, `dW=0`
So, `dQ = dU`
`Q=msdT = 5xx0.172xx10`
`=8.6 cal = 8.6 xx 4.2 J`
`36.12J`.